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Quantum Mechanics I: PHY-212
Spring Semester 2014
Problems on Wave Mechanics
1. An atom is moving in vacuum with a slow speed v ≪ c. A Gaussian wave packet
pilots the electron as shown below.
x1>x2>x3
x1
v
v
v
0
x3
x2
At some
later time
x
At some even
later time
Show that the group velocity vg is equal to the physical speed of the particle. Show that
the phase velocity is NOT the speed of the particle. Use the Einstein’s relationship
E = ~ω and the classical energy E = p2 /2m which is relevant for a slow particle.
2. Now use the relativistic expression for energy E =
√
(m0 c2 )2 + (pc)2 to calculate the
phase velocity and group velocity.
This description applies to the question 3 and 4.
As a consequence of the Heisenberg uncertainty principle the more closely an electron
is confined to a region of space the higher its kinetic energy will be. In an atom
the electrons are confined by the Coulomb potential of the nucleus. The competition
between the confining nature of the potential and the liberating tendency of the uncertainty principle gives rise to various quantum mechanical effects. Some of these
microscopic effects have repercussions in the way this universe is structured.
3. (a) Use the uncertainty principle to estimate the kinetic energy of an electron confined
within a given radius r in a hydrogen atom. Assume that ∆p ∼ p and ∆r ∼ r.
(b) Hence estimate the size of the hydrogen atom in its ground state by minimizing
it total energy as a function of the orbital radius of the electron.
(c) Compare the size obtained in this way with the value obtained from a Bohr theory
calculation.
4. When atoms are subjected to a high enough pressure they become ionized. This will
happen, for example, at the center of a sufficiently massive gravitating body.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
(a)
Spring Semester 2014
In order to ionize an atom a certain minimum energy must be supplied to
it, 13.6 eV, in the case of hydrogen.
Estimate the reduction in atomic radius re-
quired to ionize a hydrogen atom.
(b) What pressure P is needed to bring this about? (Hint: P = −dE/dV where E
is energy and V is the volume.)
(c) A planet is defined as a body in which the atoms resist the compressive force of
gravity. Estimate the maximum mass and size of a planet composed of hydrogen. (You
will need to estimate the pressure required at the center of the planet to support a
column of mass against its weight.)
This turns out to be of the order of the mass of Jupiter. Thus, Jupiter is not only the
largest planet composed of hydrogen in the solar system but anywhere in the universe!
5. The wave function of a free electron is,
10
ψ(x) = Aei(5×10
x)
,
where x is in meters. Calculate,
(a) the de Broglie wavelength,
(b) its momentum,
(c) its kinetic energy,
(d) uncertainty in position and momentum.
6. The wave function of a particle confined to move in a one-dimensional box of length
L is,
(
)
2πx
ψ(x) = A sin
·
L
(a) Use normalization condition to find the value of A.
(b) Calculate the probability of finding the electron between x = 0 to x = L/4.
7. Consider a particle whose wave function is given by the following expression,
ψ(x) = Ae−ax .
2
(a) What is the value of A if this wave function is normalized?
(b) What is the expectation value of x for this particle? What does the “expectation
value” mean?
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
8. A stationary 1 mg grain of sand is found to be at a given location within an uncertainty
of 550 nm.
(a) What is the minimum uncertainty in its velocity?
(b) Were it moving at this speed, how long would it take to travel 1 µm?
(c) Can classical mechanics be applied reliably?
(d) What is a reasonable wavelength of the grain of sand and will it behave as a wave
or as a particle?
(e) What is the minimum uncertainty in its velocity if h = 6.67 × 10−10 Js instead of
6.67 × 10−34 Js.
9. An electron is held in orbit about a proton by electrostatic attraction.
(a) Assume that an “orbiting electron wave” has the same energy an orbiting particle
would have if at radius r and of momentum mv. Write an expression for this energy.
(b) If the electron behaves as a classical particle, it must obey F = ma. Assuming
circular orbit, apply F = ma to eliminate v in favor of r in the energy expression.
(c) Suppose instead that the electron is an orbiting wave, and that the product of the
uncertainties in radius r and momentum p is governed by an uncertainty relation of
the form ∆p∆r ≈ ~. Also assume that a typical radius of this orbiting wave is roughly
equal to the uncertainty ∆r, and that a typical magnitude of the momentum is roughly
equal to the uncertainty ∆p, so that the uncertainty relation becomes pr ≈ ~. Use
this to eliminate v in favor of r in the energy expression.
(d) Sketch on the same graph the expressions from parts (b) and (c).
(e) Find the minimum possible energy for the orbiting electron wave, and the value of
r to which it corresponds.
10. A beam of electrons is accelerated through a potential difference V . The beam is
then incident on a screen with two slits and a viewing screen is placed far away. An
interference pattern is viewed on the screen as shown.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
l
I
Z +ve
d
Z=0
0
Z -ve
V +
Viewing
screen
(b) Pattern seen on screen
(a) Experimental arrangement
Z
(c) Profile of the pattern
(a) Why is an interference pattern observed?
(b) Suppose the experiment is repeated with an accelerating potential of 0.5 V instead
of V . How will the interference pattern change? Sketch the pattern clearly as well as
the profile and compare with the original observations shown in Fig.(a) and (b).
(c) If V is reduced to 0.5 V, what is the effect on:
(i) K.E. of each electron reaching the slits?
(ii) Momentum of each electron reaching the slits.
(d) If the acceleration potential is still V (with the direction of V reversed of course),
how would the interference profile change if one were to use protons instead of electrons?
11. An electron is confined in an infinite well of 30 cm width.
(a) What is the ground-state energy?
(b) In this state, what is the probability that the electron would be found within 10
cm of the left-hand wall?
(c) If the electron instead has an energy of 1.0 eV, what is the probability that it
would be found within 10 cm of the left-hand wall?
(d) For the 1-eV electron, what is the distance between nodes and the minimum
possible fractional decrease in energy?
12. A 50 eV electron is trapped in a finite well. How “far” (in eV) is it from being free
if the penetration length of its wave function into the classically forbidden region is 1
nm?
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
13. Consider a particle that is bound inside an infinite well whose “floor” is sloping as
shown.
U(x)
U(x)
oo
oo
E1
x=L
E2
x= 0
Sketch a plausible wave function when the energy is E1 and when the energy is E2 .
14. In an infinite well, consider the 1st excited state, i.e., n = 2.
(a) What is the most probable position of the particle after a measurement has been
made?
(b) What is the average position, ⟨x⟩?
15. The nuclear potential that binds protons and neutrons in the nucleus of an atom
is often approximated by a square well. Imagine a proton confined in an infinite
square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength
and energy associated with the photon that is emitted when the proton undergoes a
transition from the first excited state (n = 2) to the ground state (n = 1). In what
region of the electromagnetic spectrum does this wavelength belong?
16. Consider the potential energy barrier of length L and height V0 as shown below. An
electron is injected from the left. It has energy E < V0 .
II
I
III
Vo
E
Electron
V(x)
x=L
x=0
Fig. (a)
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
(a) Write down the wavefunctions in regions I, II and III. These wavefunctions should
include physically plausible terms. The Schrodinger equation (space part) is,
−
~2 d2 ψ(x)
+ V (x)ψ(x) = Eψ(x).
2m dx2
(b) Write down the boundary conditions at x = 0 and x = L.
(c) If a single electron is injected, will it be reflected from the wall at x = 0? Can it
penetrate through the obstacle and be found at x > L?
(d) Can the electron be “really”—I mean physically be found inside the region II? Use
the uncertainty principle to answer this question.
(e) Find the probability T that the incident electron from the far left is transmitted
into region III.
(f) Now consider the Fig.(b) with, E < V0 , E > W0 , and V0 > W0 .
II
E
I
III
Vo
Wo
x=L
x=0
Fig. (b)
Using your result for part (e), find the transmission probability into region III.
(g) If the barrier in figure (a) is to act like a 50:50 beamsplitter, what are the required
conditions on E, V0 and L?
17. The figure shows the potential energy landscape and the dashed line shows the energy
of an electron trapped inside the infinite well.
V
oo
V
oo
E
x=0
Date: 21 April, 2014
L/2
x=L
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Quantum Mechanics I: PHY-212
Spring Semester 2014
The possible wave function is:
x=L
x=0
x=L
x=0
(a)
(b)
x=0
x=L
x=0
(c)
x=L
x=0
x=L
(e)
(d)
18.
E
II
I
Vo
Vo
A particle (photon) is injected from the left into the region I, it encounters a potential
step of height V0 and enters region II. The energy of the photon is 2V0 . What can you
say about the ratio of the propagation speeds of the photon in regions I and II, vI and
vII ?
(a) vI = vII .
(b) vI < vII .
(c) vI =
(d) vI =
v√II
.
2
√
2vII .
(e) insufficient information is available to answer this question.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
√
19. A free electron of energy E has a de Broglie wavelength λ = h/p = h/ 2mE and
speed v. In the presence of an electric field, it acquires a potential energy −eu(x),
where u(x) is the potential. Hence the total energy changes, and the speed of the
electron changes to v ′ . What is the value of refractive index n =
v
?
v′
(a) 1 (one).
√
(b) E/u(x).
√
(c) E/(E − eu(x)).
√
(d) (E − eu(x))/E.
√
(e) eu(x)/E.
20. A particle is described by the wavefunction Ψ(x, t) = ei(kx−ωt) and can be thought
of a plane wave traveling along the x axis. The real part at t = 0 is shown in the
accompanying diagram. (The wavefunction extends from −∞ to ∞ which of course we
cannot show on paper.) Which of the following statements most accurately describes
the probability of finding the particle.
Re [ψ(x,t= 0)]
x
(a) It is equally likely to find the particle anywhere along the x axis.
(b) It is most likely to be found in the peaks of the wave.
(c) It is most likely to be found in the peaks or the troughs the wave.
(d) The position of the particle depends on when I make a measurement.
(e) I have no idea how to answer this question.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
21. An electron is trapped in a quantum dot of diameter L. The electron is in a potential
well of depth V0 .
Vo
L
π ~
The energy values are approximately the same as the infinite well, E = n2 2mL
2. A
2 2
laser photon of energy Ephoton shines on the quantum dot in the ground state. What
should be the minimum diameter if the electron is to always remain confined in the
quantum dot?
22. The position wavefunction, ψ(x) of a particle at some instant is given by,
ψ(x) =
L2
1
,
+ (x − xo )2 /α2
where L, xo , α are constants. Which of the following expressions given below is a good
approximate to the spread in the momentum, ∆p? We are measuring spreads by
FWHM (full width at half maximum), and ∆x∆p ≥ ~. (HINT: The wavefunction is
maximum at x = x0 ) and its profile is shown in the figure.
ψ(x)
A
A/2
xo
x
x
(a) ∆p ∼ ~/(αL)
(b) ∆p ∼ ~/α
(c) ∆p ∼ ~/(α2 xo )
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
(d) ∆p ∼ ~/(α2 L2 )
(e) ∆p ∼ ~/(αxo )
23. The energy landscape for electrons in a metal is shown below in Fig.(a), while Fig.(b)
shows the physical configuration.
(a)
(b)
φ
EF
Metal
d
Metal Tip
Vo
Metal
Vacuum
The electrons have quantized energies and the most energetic electron is at the energy
EF (called the Fermi energy). Now EF is below the vacuum level by an energy ϕ called
the work function.
(a) In field emission, a large positive voltage V0 is applied to a nearby metal tip as
shown in Fig.(b). Sketch how the energy diagram shown in (a) is modified. How are
electrons ejected from the metal?
(Don’t confuse voltage V0 with potential energy V ).
(b) Using your sketch of the modified potential energy diagram, find the minimum
voltage V0 required for the electron to tunnel out into the metal tip kept a distance d
away. The electron does not change its energy in the process.
(c) Assume that the critical distance for the tunneling of this kind is d = 2.5 nm and
the work function is 4 eV. What is the electric field required for emission?
24. Advance an argument that there is no bound (=quantized) state in a half infinite well
(shown in the diagram) unless the potential barrier V0 is at least h2 /(8ma2 ).
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
oo
Vo
x=0
V=0
x=a
25. The equations for the reflection R and transmission T for light encountering a transparent film are essentially the same as a particle of energy E seeing a potential discontinuity while E always remains greater than the potential energy. Derive the value
of the ratio between the wave numbers k1 and k2 if 50% of the light is always reflected
at an interface.
26. A metal is held at zero voltage. The energy diagram at the metal-air interface is
shown.
Unfilled levels
φ
EF
Filled levels
Metal
Air
In thermionic emission, electrons are ejected from the metal surface because:
(a) The work function ϕ increases.
(b) The work function ϕ decreases.
(c) The potential energy seen by the electrons in the air slopes downward.
(d) Increasing temperature makes more electrons jump into unfilled levels increasing
the fraction of electrons with thermal energy beyond ϕ.
(e) The Fermi level EF decreases.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
27.
n (quantum number)
10
9
8
7
6
5
4
3
2
1
i
EF
Metal
e2 /2Cdot
Quantum dot
Oxide
layer
The figure shows the energy diagram for a metal in which electrons fill energy levels
upto EF . A thin insulating oxide layer separates the metal from a quantum dot with
only ten quantized energy levels. The quantum dot is given a positive potential V0
with respect to the metal, enabling an electron to tunnel across the oxide layer. Which
one of these plots shows the correct behavior of the tunneling current i from metal to
the quantum dot. At V0 = 0, EF is at the same energy as the n = 7 quantum level.
(a)
i
(b)
i
i
(c)
e /2Cdot
Only three peaks
are observed
Vo
i
(d)
i
e /2Cdot
Only four peaks
are observed
Vo
Date: 21 April, 2014
Vo
e /2Cdot
Vo
(e)
e /2Cdot
Ten peaks are
observed
Vo
12
Quantum Mechanics I: PHY-212
Spring Semester 2014
28. An electron is injected into a potential energy landscape from the left region I as shown
below. It encounters a potential step. The energy of the electron is E and E < |V0 |.
If the electron is to emerge in region III with a faster speed, the appropriate potential
step is given by which of the following?
(a)
(b)
II
I
Vo
II
I
E
III
Vo
E
E
I
III
(c)
(d)
II
II
Vo
III
I
E
Vo
III
(e) The speed of the electron cannot incraese.
29. Snell’s law of refraction determines the bending of light across an interface. For sure,
electrons are also waves and can be refracted. The corresponding law for electrons is
called Bethe’s law and is given by,
sin α
v2
= ,
sin β
v1
where α is the angle of incidence measured from the normal to the interface, β is the
angle of refraction also measured from the normal, v1 is the speed of electron in the
incident medium and v2 is the speed in the refracted medium. Now a beam of electrons
is made to pass through two hollow cylinders with an applied voltage difference. Which
of the following diagrams show the correct trajectory of electrons?
Date: 21 April, 2014
13
Quantum Mechanics I: PHY-212
Spring Semester 2014
Electron gun
Electron gun
Incident
medium
α
+
+
+
+
+
+
β
+
+
+
+
+
+
+
+
+
+
+
+
refracted
medium
β
+
+
+
+
+
+
refracted
medium
(b) Bending towards normal.
(a) No bending.
Electron gun
Electron gun
Incident
medium
α
+
+
+
+
+
+
Incident
medium
α
β
+
+
+
+
+
+
refracted
medium
(c) Bending away from normal.
Incident
medium
+
+
+
+
+
+
+
+
+
+
+
+
refracted
medium
(d) No refraction takes place.
(e) None of these.
30. An electron starts off in the region B, trapped in a well. The potential energy V (x)
along position x is shown.
Region A
Region B
Region C
E= Energy of the electron
x
Date: 21 April, 2014
14
Quantum Mechanics I: PHY-212
Spring Semester 2014
Now suppose some time lapses. From a quantum viewpoint, which of the regions A or
C, is the electron more probable to be found?
(a) Region A.
(b) Region C.
(c) Equal probability of being found in A and C.
(d) The electron absolutely cannot leave region B.
(e) None of the above.
31. A particle moving in a region of zero force encounters a precipice—a sudden drop in
the potential energy to an extremely large negative value. What is the probability
that it will “go over the edge”, i.e., it will enter the negative potential energy region?
(a) Almost zero.
(b) Almost one.
(c) ≈ 1/2.
(d) > 1/2.
(e) None of the above.
32. An electron is trapped inside a three-dimensional quantum dot. The energy is quantized in three dimensions according to,
Enx ,ny ,nz
(
)
π 2 ~2 n2x n2y n2z
=
+ 2 + 2 ,
2m a2
b
c
where a, b and c are the confining dimensions of the box(= dot) and nx , ny , nz are
the three quantum numbers, each one of them being a positive integer.
`Quantum Dot''
''
b
c
a
If a = b = c, the energy difference between the ground and the first excited state is,
Date: 21 April, 2014
15
Quantum Mechanics I: PHY-212
(a)
Spring Semester 2014
π 2 ~2
.
2ma2
π ~
(b) 9 2ma
2.
2 2
π ~
(c) 3 2ma
2.
2 2
(d)
π 2 ~2
.
ma2
(e) There are more than one “first excited states” all with different energies. Hence
this question cannot be answered.
33. A free particle has a wave function,
Ψ(x, t) = Aei(2.5×10
11 x−2.1×1013 t)
,
where x is in metres and t is in seconds. What is the mass of the particle?
(a) Mass can only be determined if A is known.
(b) 0.012 kg.
(c) 0.11 kg.
(d) 5.7 × 10−16 kg.
(e) 1.7 × 1015 kg.
34. An electron of energy 1 eV is trapped inside an infinite well of length 30 cm. What is
the distance between two consecutive nodes of the electron’s wavefunction? (A node
is a point where the wavefunction goes to zero.)
(a) There are no nodes in the electron’s wavefunction.
(b) The distance between consecutive nodes is zero.
(c) 1.25 × 10−18 m.
(d) 6 × 10−10 m.
(e) None of the above.
35. An electron is trapped in an infinite well of length L and ground state energy E1 . At
t = 0, the wavefunction is,
)
1 (
ψ1 (x) + 2ψ2 (x) ,
Ψ(x, 0) = √
5L
where ψ1 (x) and ψ2 (x) are normalized wavefunctions in the ground and first excited
states. The wavefunction at t = π~/E1 is given by:
Date: 21 April, 2014
16
Quantum Mechanics I: PHY-212
Spring Semester 2014
(a)
(b)
Re (ψ)
Re (ψ)
1
2
0.2
0.4
0.6
0.8
1.0
x/L
1
-1
0.2
0.4
0.6
0.8
1.0
x/L
-2
-1
(d)
(c)
Re (ψ)
Re (ψ)
1
2
0.2
0.4
0.6
0.8
1.0
x/L
1
-1
0.2
0.4
0.6
0.8
1.0
x/L
-2
-1
Re (ψ)
(e)
4
3
2
1
0.2
0.4
0.6
0.8
x/L
1.0
36. The potential energy profile in a certain region is shown.
to oo
to oo
energy E
a
b
x
A particle of energy E exists inside this region. A sketch of the possible (real part) of
the wavefunction is;
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
(b)
(a)
wavefunction is
non zero at x > b
wavefunction
is zero at x > b
b
a
x
(d)
(c)
b
a
x
b
a
wavefunction
is zero at x > b
wavefunction
is zero at x > b
x
a
x
b
(e) None of the above.
37. Suppose a particle is in the ground state with wavefunction ψ1 (x). Which one of the
following is the probability that the particle will be found in a narrow range between
x and x + dx.
(a) |ψ1 (x)|2 dx.
(c)
∫dx
x|ψ1 (x)|2 dx.
x
(b) x|ψ1 (x)|2 dx.
(d)
+∞
∫
−∞
x|ψ1 (x)|2 dx.
(e) None of the above.
38. A free particle has a wavefunction A(eikx + e−ikx ) and energy E. A is a normalization
constant. Mark True of False against these statements.
(i) The probability density does not change with time.
(ii) The probability density is constant in space x.
(iii) The de Broglie wave associated with the particle is in fact a standing wave.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
39. The uncertainty relationship for a particle moving in a straight line is ∆p∆x ≥ ~/2.
S
R
θ
R = radius
S = Arc length
If the particle is moving in a circle with angular momentum L, the uncertainty relationship becomes:
(HINT: Distance becomes the arc length!)
(a) ∆L∆θ ≥ ~2 .
(b) ∆L∆S ≥ ~2 .
(c) ∆L∆R ≥ ~2 .
(d) ∆L∆θ ≤ ~2 .
(e) None of the above.
40. In a scanning tunneling microscope (STM), the tunneling probability of electrons from
metal surface to a prob tip is proportional to exp(−2αL), where L is the tip-sample
distance and α = 1 nm−1 is the inverse of the penetration length.
Tip
L
Metal surface
Tip
L -∆L
Metal surface
If the tip moves closer to the surface by ∆L = 0.1 nm, the tunneling current,
(a) remains unchanged.
(b) increase by 22 %.
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
(c) decrease by 22 %.
(d) increase by 10 %.
(e) decrease by 10 %.
41. The Heisenberg uncertainty principle applies to photons as well as to material particles.
Thus a photon confined to a small box of size ∆x necessarily has a large uncertainty
in momentum and uncertainty in energy. Recall that for a photon E = pc.
(a) Estimate the uncertainty in energy for a photon confined to the tiny box of size
∆x.
(b) If ∆E ∼ E, what is the effective mass of the photon?
(c) This mass can be extremely large, if ∆x is tiny. If ∆x is sufficiently small, the large
mass can create a large gravitational field, sufficiently large to form a black hole. When
this happens ∆x is called the Planck length, and this is when gravity and quantum
mechanics become inter mixed. For a black hole, not even light can escape.
Launch an object
M
R
Consider a star of mass M and radius R as shown above. If an object is to be
launched from the star’s surface so that it escapes the star’s gravitational pull, it
√
needs a minimum velocity vesc called the escape velocity. Show that vesc = 2GM
.
R
(d) If vesc = c, nothing can escape from this star, not even light. If we were to replace
the star of mass M with a photon of the mass calculated in part (b), and confined to
length ∆x, and set R = ∆x, calculate the Planck length in terms of G, ~ and c.
(e) If G = 6.67 × 10−11 N m2 /kg2 , find the numerical value of Planck’s length.
(f) What is the diameter of a proton (about 2 fm = 2 × 10−15 m) in units of Planck’s
length?
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
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42. The radioactive decay of certain heavy nuclei by emission of an alpha particle is a result
of quantum tunneling. Imagine an alpha particle moving around inside a nucleus,
such as thorium (mass number= 232). When the alpha particles bounces against the
surface of the nucleus, it meets a barrier caused by the attractive nuclear force. The
dimensions of barrier vary a lot from one nucleus to another, but as representative
numbers you can assume that the barrier’s width is L ≈ 35 fm (1 fm = 10−15 m) and
the average barrier height is such that V0 − E ≈ 5 MeV. Find the probability that an
alpha hitting the nucleus surface will escape. Given that the alpha hits the nuclear
surface about 5 × 1021 times per second, what is the probability that it will escape in
√
a day? The tunneling probability is T = e−2αL where α = 2m(V0 − E)/~ and L is
the barrier length. (1 MeV= 106 eV).
Date: 21 April, 2014
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Quantum Mechanics I: PHY-212
Spring Semester 2014
Solution
Answer 1:
The group velocity is given by,
dω
vg =
=
dk
(
dω
dE
)(
)
dE
.
dk
(1)
According to Einstein’s energy equation,
E = ~ω.
(2)
Classically, the energy of a slow moving particle is given by,
1 2
p2
E = mv =
,
2
2m
(3)
where p = momentum of the particle = h/λ = (h/2π)(2π/λ) = ~k. Slow means v ≪ c,
implying that we can use the non-relativistic expressions. Therefore equation (38) becomes,
E=
~2 k 2
.
2m
(4)
From equation (37),
dE
= ~
dω
dω
1
⇒
= ,
dE
~
(5)
while from equation (39) we obtain,
dE
2~2 k
~2 k
=
=
.
dk
2m
m
(6)
Putting equations (40) and (41) into equation (42),
vg =
1 ~2 k
~k
p
=
=
= vparticle ,
~ m
m
m
as required.
Whereas the phase velocity is given by,
vp =
ω
E~
E
p2 /2m
p
vparticle
=
=
=
=
=
̸= vparticle .
k
~p
p
p
2m
2
This clearly shows that for the atom, vparticle is exactly reproduced by the group velocity of
the associated matter wave, whereas the phase velocity is half the particle velocity.
Date: 21 April, 2014
22
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 2:
For convenience, let’s denote vparticle = v. The energy and momentum of a relativistic
particle (in this case, electron) are given by,
mo c 2
E = m c2 = √
·
2
1 − vc2
mo v
p = mv = √
·
v2
1 − c2
(7)
The energy in terms of the relativistic momentum p and the rest mass mo can be obtained
from the expression we have now encountered several times,
E 2 = p2 c2 + m2o c4
√
E = c p2 + m2o c2 .
(8)
The corpuscular features (energy and momentum) of an electron are connected to its wave
characteristics (wave frequency and number ) by the relations,
E = ~ω
p = ~ k.
and
Therefore the group and phase velocities will become,
dω
dE
=
,
dk
dp
E
E
=
=
·
p
p
vg =
vph
and
From Equation (23) and (24), we find that p2 + m2o c2 =
m2o c2
2
(1− v2 )
. So, the phase velocity is
c
vg
( √
)
dE
d
2
2
=
=
c p + mo c
dp
dp
pc
= √
p2 + m2o c2
√
2
mo v c/ 1 − vc2
=
E
c
mo v c 2
√
=
2
E 1 − vc2
mo v c 2
= √
×
2
1 − vc2
√
1−
v2
c2
mo c 2
·
Hence vg = v.
Date: 21 April, 2014
23
Quantum Mechanics I: PHY-212
Spring Semester 2014
This shows that the speed of relativistic particle is equal to its group velocity.
Similarly, the phase velocity of relativistic particle can be calculated as
vph =
=
=
=
=
=
Hence vph =
E
p
√
c p2 + m2o c2
p
√
m2 c 2
c 1 + o2
p
√
m2 c2
v2
c 1 + 2o 2 × (1 − 2 )
mo v
c
√
c2
v2
c 1 + 2 (1 − 2 )
v
c
√
2
c
c
.
2
( v)
c
c
.
v
As c > v, this means vph > c, predicting that the phase velocity for the relativistic particle is
greater than the speed of light c. This appears to be a violation of the postulates of special
theory of relativity. Actually, the phase velocity does not represent the physical velocity of
the particle, rather it is the group velocity which represents the speed of propagation of the
particle. Hence, the result that vg , and not vp , represents particle speeds, holds both in the
relativistic and non-relativistic scenarios.
Answer 3:
(a) It is given that ∆p ∼ p and ∆r ∼ r. When we consider small radii, the electron is
present very close to the nucleus. Pushing the electron any closer to the nucleus results in
increased energies. The electron may even gain enough energy to fly away from the nucleus.
This is when the atom will ionize and hence the useful rule, “it is impossible to squish
atoms”. Close to the nucleus, we are “rubbing shoulders” with the uncertainty principle.
According to this principle, the momentum of an electron confined within a given radius
r is approximately given by p ∼ ~/r. (One could also use p ∼ ~/(2r) without affecting the
overall implications of the result. Remember that the uncertainty principle is an inequality!)
Therefore, when confined to a radius r, the kinetic energy will be of the order,
K.E =
Date: 21 April, 2014
p2
~2
=
.
2m
2mr2
24
Quantum Mechanics I: PHY-212
Spring Semester 2014
Attempting to bring the nucleus any closer to the nucleus may result in extremely large
kinetic energies, shooting the electron away.
(b) In the closest approach of the electron to the nucleus, the total energy of the hydrogen
atom is,
Total Energy = E = K.E + P.E
~2
Ze2
=
·
−
2mr2 4πϵo r
The energy is minimum when dE/dr = 0,
dE
~2 d 1
Ze2 d 1
=
( 2) −
( )
dr
2m dr r
4πϵo dr r
~2 −2
Ze2 −1
=
( 3 )−
( )
2m r
4πϵo r2
Ze2
−~2
=
+
·
mr3
4πϵo r2
Setting this equal to zero,
−~2
Ze2
+
= 0
3
2
mrmin
4πϵo rmin
Ze2
~2
=
2
3
4πϵo rmin
mrmin
4πϵo ~2
rmin =
·
mZe2
= 0.53 Å,
Energy (eV)
30
The uncertainty principle prevents the electron
from approaching the nucleus at short distances
20
10
r min
Radius
-10
E min
This is the radius, rmin , when the energy is minimum. The nucleus attracts the electron,
so the electron prefers to exist close to the nucleus, but at the same time, the uncertainty
principle does not let it come too close!
Date: 21 April, 2014
25
Quantum Mechanics I: PHY-212
Spring Semester 2014
(c) The value of the radius calculated above is in excellent agreement with the radius of the
smallest orbit (n = 1) calculated from Bohr’s model.
Answer 4:
(a) Using the information provided in Question 1: ∆p ∼ p and ∆r ∼ r , and using the uncertainty principle, the momentum of an electron confined within a radius r is approximately
p ∼ ~/r. The total energy is,
Total Energy = E = K.E + P.E
~2
e2
=
−
.
2mr2 4πϵo r
(9)
Ionization occurs when the energy of the electron approached zero, the energy of the vacuum
state. We calculate the radius rion when E = 0.
~2
e2
−
= 0
2mr2 4πϵo r
~2
4πϵo r
=
2
2mr
e2
2πϵo ~2
= 0.24 Å.
rion =
me2
The radius rion is smaller than the rmin calculated from the previous question, as we expect.
Excessive pressure inside a planet can push the electron to this radius. At this point, the
atoms will ionize and the planet will not be stable.
(b) The pressure is given as,
dE
dV
dE dr
= −
dr dV
P = −
using the chain rule.
Furthermore, we have,
4 3
πr
3
dV = 4πr2 dr
dr
1
=
.
dV
4πr2
V =
Differentiating the energy expression from (9),
( )
( )
dE
~2
1
−2
2 −1
=
−
e
dr
2m r3
4πϵo
r2
−~2
e2
=
+
.
mr3 4πϵo r2
Date: 21 April, 2014
26
Quantum Mechanics I: PHY-212
Spring Semester 2014
We now substitute the value of the radius, r = rion ,
)3
(
)2
(
e2
1
dE −~2 1
+
= −3.9 × 107 J m−1 ,
=
dr r=rion
m rion
4πϵo rion
resulting in the ionizing pressure,
dE
1
2
dr 4πrion
= 5.2 × 1013 Pa.
Pion = −
(c)
R
Pion
We assume a spherical planet of radius R and mass M . We determine the parameters that
result in ionizing pressures at the centre of the planet. First of all, we assume a constant
density ρ of the planet throughout the interior. An estimate of the density is the proton
mass divided by the volume of the atom,
ρ=
mp
4
3
πrion
3
= 2.8 × 104 kg m−3 .
(10)
The pressure exerted by a fluid of length R at its base is given by ρgR. However, the
value of g on this planet is unknown, but from Newton’s law of gravitation, we know that
g = GM/R2 . Therefore,
Pion = ρ g R =
⇒R=
ρGM
R
(11)
ρGM
= 3.5 × 10−20 M m.
Pion
(12)
Now the density ρ can also be equated to the mass of the planet divided by its volume,
ρ = 2.8 × 104 kg m−3 =
⇒M =
Date: 21 April, 2014
M
4
πR3
3
4
πρR3 .
3
(13)
(14)
27
Quantum Mechanics I: PHY-212
Spring Semester 2014
Inserting the value of M into (12) and then back substituting results in,
M = 4 × 1026 kg,
R = 1.6 × 107 m.
The measured mass and radius of Jupiter are 1.9 × 1027 kg and 7 × 107 m (values taken from
Wikipedia).
Answer 8:
We are given that,
Mass of grain = m = 1 mg = 10−6 kg
Uncertainty in position = ∆x = 550 nm = 550 × 10−9 m.
(a) Uncertainty in velocity can be calculated by calculating uncertainty in its momentum.
According to uncertainty principle, the minimum uncertainty is approximately,
∆x ∆p ≥
~
2
~
1.05 × 10−34
=
2∆x
2 × 550 × 10−9 m
= 9.65 × 10−29 kgms−1 .
⇒ ∆p ≥
∆p is small because ~ is small. Now the uncertainty in speed is calculated as,
∆p = m∆v
∆p
∆v =
m
9.7 × 10−29 kg m s−1
=
10−6 kg
= 9.65 × 10−23 m s−1 .
For macroscopic particle ∆v ≥ ~/2(∆x)m is small because of the very small ~/m ratio. ∆v
becomes significant only if ~ were large or the mass m decreases. Small ~ and large m makes
the macroscopic classic world “undisturbed” by quantum uncertainties!
(b)
∆t ≈
1 µm
10−6
=
s = 0.1 × 1017 s ≈ 3 billion years!
∆v
9.65 × 10−23
Date: 21 April, 2014
28
Quantum Mechanics I: PHY-212
Spring Semester 2014
The uncertainty in velocity is really really small! An observer would require 3 billion years
to notice the grain of sand, supposedly at rest, at a position 1 µm away from its original
position. The current age of the solar system is approximately 5 billion years.
(c) Yes uncertainties are extremely small. No device has ever been built, and may never be
built that can measure these small velocities. We can safely apply classical mechanics to a
grain of sand; there is effectively no uncertainty in position or in momentum. Furthermore,
a precision as fine as 10−22 m/s is never required in classical mechanics.
(d)
λ=
h
h
= .
mv
p
Now what momentum should I choose? The uncertainty principle dictates a ∆p ∼ 9.7×10−29
kg m s−1 . The momentum could therefore have any value between, approximately −∆p/2
and ∆p/2. Let’s choose an extreme value, p ∼ ∆p/2 ∼ 5 × 10−29 kg m s−1 . Therefore,
λ∼
6.67 × 10−34
∼ 1.3 × 10−5 m.
5 × 10−29
This is such a small wavelength compared to apparatus we might use for macroscopic objects,
that for all practical purposes, the grain of sand acts like a particle!
(e) ∆v would be 9.65 × 10 ≈ 96 m/s, if h were this large. This is a huge uncertainty. We
are “saved” by the exceedingly small value of h
Answer 9:
(a) Total energy of an orbiting particle in terms of its kinetic energy and electrostatic
potential energy is give by,
Etotal = K.E. + P.E.
[
]
1 2
ke2
Etotal = mv + −
2
r
2
ke
1
,
Etotal = mv 2 −
2
r
where k = 1/4πϵ0 . Therefore total energy of the particle will become,
Etotal =
1 2
e2
mv −
·
2
4πϵ0 r
Hence the energy of “orbiting electron wave” is also,
1
e2
E = mv 2 −
·
2
4πϵ0 r
Date: 21 April, 2014
(15)
29
Quantum Mechanics I: PHY-212
Spring Semester 2014
(b) Since electron is orbiting in a circular orbit, its centripetal acceleration in its orbit is
(v 2 /r), while electrostatic force on the electron is (ke2 /r2 ), thus,
F = ma
( 2)
ke2
v
= m
2
r
r
2
ke
⇒ v2 =
mr
e2
2
v =
.
4πϵ0 mr
Use this value of v 2 in equation (42),
Eclassical particle =
=
=
=
(
)
1
e2
e2
m
−
2
4πϵ0 mr
4πϵ0 r
( 2 )
2
1
e
e
−
2 4πϵ0 r
4πϵ0 r
2
2
e
e
−
8πϵ0 r 4πϵ0 r
e2
−
·
8πϵ0 r
The negative electrostatic potential energy is always of greater magnitude than the positive
kinetic energy, so the total energy strictly decreases as r decreases. Hence there is no
minimum energy. In the accompanying figure, course A corresponds to the energy of the
classical particle, whose energy decreases as r.
(c) Now assuming pr = ~, we have p = ~/r or v = ~/mr. Therefor equation (42) becomes,
Ematter wave
(
)2
1
~
e2
= m
−
2
mr
4πϵ0 r
2
2
~
e
=
−
·
2
2mr
4πϵ0 r
In this case as r decreases, and the wave become more compact, the likely speed increases.
The kinetic energy increases faster than the potential decreases, and the total energy at
some point must increase. Hence applying uncertainty principle there is a turning point A
in the curve labelled B.
(d) The two plots are shown in the figure.
Date: 21 April, 2014
30
Quantum Mechanics I: PHY-212
E
Spring Semester 2014
0.2
(arb. units)
2
4
6
8
10
r
Curve B
-0.2
A
Curve A
-0.4
-0.6
While the energy of a classical particle would monotonically decrease as r decreases, the
energy of the matter wave reaches a minimum, and then increases.
(e) The minimum possible energy for the orbiting electron wave can be calculated by setting
the derivative of energy with respect to r, to zero.
Ematter wave =
dEmatter wave
=
dr
⇒ r =
=
=
~2
e2
−
2mr2 4πϵ0 r
~2
e2
− 3+
=0
mr
4πϵ0 r2
4πϵ0 ~2
me2
9 × 109 Nm2 /C2 × (1.055 × 10−34 Js)2
9.11 × 10−31 × (1.6 × 10−19 C)2
5.3 × 10−11 m.
This turns out to be astoundingly close to the Bohr radius calculated earlier in class. Inserting this value of r and other constants will give energy for matter wave as follows.
Ematter wave = −13.6 eV.
The energy happens to equal the correct, experimentally determined value, and the radius
is indeed the most probable radius at which the electron would be found. That these
agree so closely is an accident; many approximations have been made. Nevertheless, the
uncertainty principle does impose a lower limit on the energy, and it is no accident that the
value we obtained is of the correct order of magnitude.
Answer 10:
(a) The electrons have a quantum field ψ as they are present in the region between the
Date: 21 April, 2014
31
Quantum Mechanics I: PHY-212
Spring Semester 2014
screens. This field is a superposition of two fields, ψ1 and ψ2 corresponding to two electrons
ejecting from either of the slits. The phase difference between ψ1 and ψ2 results in these
field interfering causing an interference pattern.
N.B. I don’t like the expression “wavefunction splitting” or “electron splitting”. “Electrons
interfering with themselves” is fine.
√
(b) V is halved, so energy eV = p2 /2m is halved. Hence p2 is halved or p is reduced 2
√
√
times. Since p = h/λ, λ increases 2 times and k = 2π/λ decreases 2 times. If the
√
voltage were V gave a wavenumber k, then a voltage V /2 gives a wavenumber k/ 2. The
interference pattern is proportional to cos2 (kd sin θ/2). Hence after the halving, the pattern
√
is proportional to cos2 ((kd sin θ)/2 2).
Let (kd sin θ)/2 = α. With voltage V , intensity pattern ∝ cos2 (α). With voltage V /2,
√
intensity pattern ∝ cos2 (α/ 2) = cos2 (0.707α).
Now a minimum is observed when cos2 (α) = 0
⇒ α = π/2, when voltage is V . Whereas,
⇒ 0.707α = π/2
if the voltage is V /2, the minimum appears at cos2 (0.707α) = 0
⇒
α = π/2 × 0.707, which is larger than the previous case. Hence the fringe width increases
√
2 = 1.411 times.
b
Voltage V
0
Z
2b
Voltage V/2
0
Z
(c)
(i) E = p2 /2m = eV , if V is halved E is halved.
√
(ii) If V is halved p2 is halved, so p is reduced 2 times.
(d) The protons gain the same energy as electrons, because they carry the same charge. But
protons are heavier than electrons, so their momentum is larger (p2 = 2mE), hence λ is
Date: 21 April, 2014
32
Quantum Mechanics I: PHY-212
Spring Semester 2014
shorter and k = 2π/λ is larger. Since energy pattern ∝ cos2 ((kd sin θ)/2), a large k would
result in rapid spatial variation in bright and dark fringes which will therefore be squeezed
close together.
With electrons
With protons
(not to scale)
Answer 11:
(a) For an infinite square well, the energy is,
En =
π 2 ~2 n 2
·
2mL2
For the ground state, n = 1 and the corresponding energy is,
π 2 ~2
2mL2
π 2 (1.054 × 10−34 J sec)2
=
2(9.1 × 10−31 kg)(0.3 m)2
= 6.71 × 10−37 J.
E1 =
(b) The wavefunction for an infinite square well is,
√
(
)
2
nπx
ψn =
sin
L
L
√
( )
2
πx
ψ1 =
sin
.
L
L
Date: 21 April, 2014
33
Quantum Mechanics I: PHY-212
Spring Semester 2014
The probability of finding the electron within 10 cm of the left-hand wall is,
∫ 0.1
P (0 < x < 0.1 m) =
ψ ∗ (x)ψ(x)dx
∫0 0.1
=
|ψ(x)|2 dx
0
∫
2 0.1 2
=
sin (πx/L)dx
L 0
∫
2 0.1 (1 − cos(2πx/L))
=
dx
L 0
2
∫
1 0.1
=
(1 − cos(2πx/L))dx
L 0
[
]0.1
sin(2πx/L)
1
1−
=
L
2π/L
0
= 0.21.
(c) If the electron has 1.0 eV of energy, then,
π 2 ~2 n 2
= 1.6 × 10−19 J
2
2mL
2mL2 (1.6 × 10−19 J)
n2 =
π 2 ~2
2(9.11 × 10−31 kg)(0.3 m)2 (1.6 × 10−19 J)
=
π 2 (1.054 × 10−34 J sec)2
= 2.38 × 1017
n = 4.88 × 108 .
With this energy of electron, the probability of finding it within 10 cm of left-hand wall is,
∫ 0.1
P (0 < x < 0.1 m) =
ψ ∗ (x)ψ(x)dx
0
∫
2 0.1 2
sin (nπx/L)dx
=
L 0
∫
2 0.1 (1 − cos(2nπx/L))
=
dx
L 0
2
∫
1 0.1
=
(1 − cos(2nπx/L))dx
L 0
[
]0.1
1
sin(2nπx/L)
=
1−
L
2nπ/L
0
= 0.33.
Date: 21 April, 2014
34
Quantum Mechanics I: PHY-212
Spring Semester 2014
(d) We know that,
λ
2
2L
·
λ =
n
L = n
Now the distance between the nodes is,
λ
L
0.3
=
=
2
2
4.8 × 108
= 6.15 Å.
The maximum possible fractional decrease in energy is thus,
∆E
En − En−1
=
E
En
2
n − (n − 1)2
=
n2
2
1
=
− ,
n n2
since n = 4.8 × 108 , the minimum fractional decrease in energy is,
∆E
= 4.1 × 10−9 J.
E
Answer 12:
E
δ
x= 0
Uo
x= L
The penetration depth δ is given by,
~
= 1 × 10−9 m
δ = √
2m(U0 − E)
~2
2m(U0 − E) =
(1 × 10−9 m)2
~2
U0 − E =
2m(1 × 10−9 m)2
= 38.2 meV.
Date: 21 April, 2014
35
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 13:
The plausible wavefunctions are shown in the Figure.
U(x)
oo
U(x)
oo
ψ1
node
E1
node
x=L
ψ2
E2
x= 0
Penetration into the
classically forbidden
region
The wavefunction ψ1 (x) corresponds to√energy E1 and ψ2 (x) corresponds to energy E2 . If
( i 2m(E−U (x)) )
E > U (x), the wavefunction is ∝ exp
, which is oscillatory. Larger the value
~
√
(x))
and shorter the wavelength. If E < U (x),
of E − U (x), higher the value of k = 2m(E−U
~
the wavefunction decays (damps). Note that the wavelength for ψ1 (x) is not uniform,
rather the wavelength increases, k decreases as E −U (x) decreases in going from left to right.
Answer 14:
The wavefunction for an infinite well is,
√
(
)
nπx
2
sin
L
L
√
(
)
2πx
2
sin
L
L
ψn (x) =
For first excited state, n = 2,
ψ2 (x) =
and the probability is,
P2 (x) = |ψ2 (x)∗ ψ2 (x)|
(
)
2
2 2πx
=
sin
·
L
L
To find the most probable position, we have to maximize P2 (x).
( ) (
) (
)( )
dP2 (x)
2
2πx
2πx
2π
= 2
sin
cos
dx
L
L
L
L
( ) (
) (
)
4
2πx
2πx
= 2π
sin
cos
2
L
L
L
Date: 21 April, 2014
36
Quantum Mechanics I: PHY-212
The quantity
dP2 (x)
dx
Spring Semester 2014
= 0 when x = 0, L/4, L/2, 3L/4, L but when x = 0, L/2, L, the wave-
function ψ2 (x) = 0.
Thus at x = 0, L/2, L, the probability of finding the particle is zero. The most probable
positions are x = L/4, and x = 3L/4.
(b) The average position is given by the following,
∫ L
⟨x⟩ =
x ψ2∗ (x)ψ2 (x)dx
0
∫ L
=
x |ψ2 (x)|2 dx
0
∫
2 L
=
x sin2 (2πx/L)dx
L 0
∫ L
2
x (1 − cos(4πx/L))dx
=
2L 0
∫
∫
1 L
1 L
=
xdx −
x cos(4πx/L)dx
L 0
L 0
L
=
−0
2
L
= ·
2
Answer 15:
In a square well, the energy that corresponds to n’th energy level is,
En =
π 2 ~2 n 2
·
2mL2
When a proton undergoes a transition from the first excited state (n = 2) to the ground
state (n = 1), the energy of emitted photon is,
π 2 ~2 2
(2 − 12 )
2mL2
3π 2 ~2
=
2mL2
3π 2 (1.054 × 10−34 Js)2
=
2(1.67 × 10−27 kg)(10−14 m)2
= 9.8 × 10−15 J.
∆E2→1 =
The wavelength of emitted photon is,
hc
∆E
(6.63 × 10−34 Js)(3 × 108 m/s)
=
9.8 × 10−15 J
= 2 × 10−11 m.
λ =
Date: 21 April, 2014
37
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 16:
(a) The time independent Schrodinger equation is,
−
~2 d2 ψ(x)
+ V (x)ψ(x) = Eψ(x).
2m dx2
For Region I:
For region I the Schrodinger equation, with V = 0, is,
~2 d2 ψ(x)
−
2m dx2
2
d ψ(x) 2mE
+ 2 ψ(x)
2
~)
( dx2
2mE
d
+ 2 ψ(x)
dx2
~
(D2 + k 2 )ψ(x)
where k =
= Eψ(x)
= 0
= 0
= 0,
(16)
2mE
.
~2
⇒ D2 = −k 2
D = ±ik,
which leads to the solution,
ψ(x) = Aeikx + Be−ikx .
the first term on the right is a forward propagating wave and the second term is a backward
propagating wave.
For Region II: V = V0 and E < V0
The Schrodinger equation for this region becomes,
~2 d2 ψ(x)
+ (V0 − E)ψ(x)
−
2
( 2m 2 dx2
)
~ d
−
+ (V0 − E) ψ(x)
2m dx2
( 2
)
d
2m
− 2 (V0 − E) ψ(x)
dx2
~
(D2 − α2 )ψ(x)
= 0
= 0
= 0
= 0
D = ±α,
where α2 =
2m(V0 −E)
,
~2
Date: 21 April, 2014
is a positive constant.
38
Quantum Mechanics I: PHY-212
Spring Semester 2014
Hence wave function for region II is,
ψII (x) = Ceαx + De−αx
(17)
Since the exponents α are real, ψII (x) does not represent an oscillatory function, rather it
is a decreasing function.
For Region III: V=0
Since the wavevector of region III is same as that of region I, hence the solution becomes,
ψ(x) = Eeikx + F e−ikx .
Since the wave does not seen any obstacle in region III, it cannot be reflected implying
F = 0. The wavefunction therefore, is,
ψIII (x) = Eeikx .
(b) The boundary condition at the edges of the barrier are that ψ(x) and
dψ(x)
dx
must be
continuous at both edges.
At x=0:
ψI (x = 0) = ψII (x = 0)
⇒A+B = C +D
dψI dψII and
=
dx x=0
dx x=0
ikA − ikB = αC − αD
ik(A − B) = α(C − D)
α
A − B = −i (C − D).
k
(18)
(19)
At x=L:
ψII (x = L) = ψIII (x = L)
⇒ CeαL + De−αL = EeikL
dψIII dψII =
and
dx x=L
dx x=L
αCekL − αDe−αL = ikEeikL
k eikL
CeαL − De−αL = i E kL ·
α e
(20)
(21)
Equations (37), (38), (39) and (40) are the required boundary conditions.
Date: 21 April, 2014
39
Quantum Mechanics I: PHY-212
Spring Semester 2014
(c) If a single electron is injected with energy less than the barrier height (E < V0 ), classically
the particle cannot penetrate through the barrier, it will be reflected completely. However
quantum mechanics tell us that the particle has certain probability to go through the barrier,
and this is when the transmission coefficient is non-zero.
(d) In region II the total energy is less than the potential energy, which means that the
particle appears to possess negative kinetic energy here. From equation (17) on page 3, 1/α
represents a penetrating length δ,
δ =
1
~
,
=√
α
2m(V0 − E)
which characterizes a realm of possibilities of position ∆x. Now using the uncertainty
principle ∆x∆p ∼ ~
∆p ∼
~
√
2m(V0 − E) √
= 2m(V0 − E) ∼ p.
~
The momentum p corresponds to an energy
p2
2m
= V0 − E and if we assume that the uncer-
tainty in energy ∆E is of the same order as the energy, the ∆E ∼ V0 − E.
As you immediately recognize, V0 − E is of the same order as the energy gap that localizes
the particle to the region δ. If the uncertainty is as large as the gap, there is no guarantee
that the particle can be localized.
(e) Transmission probability T is given by,
T =
=
=
=
=
Prob. that the particle crosses right boundary per unit time
Prob. that the particle crosses left boundary per unit time
|prob/time|x=L
|prob/time|x=0
|prob/length × length/time|x=L
|prob/length × length/time|x=0
|ψ(x)|2x=L × v
|ψ(x)|2x=0 × v
|ψ(x)|2x=L
.
|ψ(x)|2x=0
In the given case, |ψ(x)|2x=L = |E|2 and |ψ(x)|2x=0 = |A|2 . therefore transmission probability
is,
T =
Date: 21 April, 2014
|E|2
·
|A|2
40
Quantum Mechanics I: PHY-212
Spring Semester 2014
In order to calculate |E|2 and |A|2 we will use the four boundary conditions derived in part
(b). Add equation (37) and (38),
(
)
(
)
α
α
2A = C 1 − i
+D 1+i
k
k
(
)
(
)
α
D
α
C
A =
1−i
+
1+i
.
2
k
2
k
(22)
Next adding equation (39) and (40) we obtain,
)
(
k
2Ce = Ee
1+i
α
)
(
ikL
Ee
k
C =
.
1+i
2eαL
α
αL
ikL
Now subtracting equation (40) from (39) results in,
)
(
k
−αL
ikL
2De
= Ee
1−i
α
(
)
ikL
Ee
k
D =
1−i
.
2e−αL
α
(23)
(24)
Using values of C and D from equations (23) and (24) in equation (41) yields,
(
)(
)(
)
(
)(
)(
)
1 EeikL
k
α
1 EeikL
k
α
A =
1+i
1−i
+
1−i
1+i
2 2eαL
α
k
2 2e−αL
α
k
[
(
)
(
)]
EeikL −αL
k
α
α
k
=
e
1 + i − i + 1 + eαL 1 + i − i + 1
4
α
k
k
α
[
(
)
(
)]
ikL
2
2
2
Ee
(k − α )
(k − α2 )
−αL
αL
=
e
2+i
+e
2−i
,
4
kα
kα
and after some further rearrangement,
[
]
EeikL
(α2 − k 2 ) αL
αL
−αL
−αL
A =
2(e + e
)+i
(e − e
)
4
kα
[( αL
)
(
)]
(α2 − k 2 ) eαL − e−αL
e + e−αL
ikL
+i
.
= Ee
2
2kα
2
Using cosh x =
ex +e−x
2
−x
and sinh x = e −e
,
2
)
(
(α2 − k 2 )
ikL
A = Ee
sinh(αL) .
cosh(αL) + i
2kα
x
Now taking the complex conjugate of equation (25),
)
(
(α2 − k 2 )
∗
∗ −ikL
A = E e
sinh(αL) ,
cosh(αL) − i
2kα
Date: 21 April, 2014
(25)
(26)
41
Quantum Mechanics I: PHY-212
Spring Semester 2014
and finally multiplying equations (25) and (26) yields,
(
)
(
)
(α2 − k 2 )
(α2 − k 2 )
∗
∗ −ikL
ikL
AA = E e
cosh(αL) − i
sinh(αL) · Ee
cosh(αL) + i
sinh(αL)
2kα
2kα
(
)
(α2 − k 2 )2
2
2
2
2
|A| = |E| cosh (αL) +
sinh (αL)
4k 2 α2
(
)
(α2 − k 2 )2
|A|2
2
2
= 1 + sinh (αL) +
sinh (αL)
|E|2
4k 2 α2
(
)
(α2 − k 2 )2
= 1+ 1+
sinh2 (αL)
4k 2 α2
(α2 + k 2 )2
= 1+
sinh2 (αL)·
2
2
4k α
Substitute values of k and α we obtain the desired result.
(
)
√
( 2m(V~02−E) + 2mE
)2
|A|2
2 L
−1
~2
= T = 1 + 2m(V0 −E) 2mE sinh
2m(V0 − E)
|E|2
~
4( ~2
× ~2 )
)
(
√
V02
2 L
−1
T
= 1+
sinh
2m(V0 − E)
4E(V0 − E)
~
[
)]−1
(
√
V02
2 L
T = 1+
2m(V0 − E)
·
sinh
4E(V0 − E)
~
Transmissivity
0.8
0.6
0.4
0.2
-18
1
2
3
4
Energy X10 J
Shown is a graph of T versus E for V0 = 10 × 1.6 × 10−19 J = 10 eV and E in the range of
0 to 30 × 1.6 × 10−19 J (= 30 eV). The length of the barrier L is chosen as 1Å. Clearly as E
increases, the transmission T goes up.
Date: 21 April, 2014
42
Quantum Mechanics I: PHY-212
Spring Semester 2014
(f ) Wave functions for the three regions are,
√
2mE
ψI (x) = Ae + Be
, where k =
√~
2m(V0 − E)
ψII (x) = Ceαx + De−αx , where α =
~
√
2m(E
−
W
)
0
ψI (x) = Eeiβx , where β =
~
−ikx
ikx
The boundary conditions at x = 0 are,
ψI (0) = ψII (0)
A+B = C +D
′
and ψI′ (0) = ψII
(0)
α
A − B = −i (C − D),
k
(27)
(28)
while the boundary conditions at x = L are,
ψII (L) = ψIII (L)
CeαL + De−αL = EeiβL
′
′
ψII
(L) = ψIII
(L)
β
Ceαx − De−αL = i EeiβL .
α
(29)
and
(30)
Adding equations (27) and (28),
(
)
(
)
α
α
2A = C 1 − i
+D 1+i
k
k
(
)
(
)
C
α
D
α
A =
1−i
+
1+i
,
2
k
2
k
(31)
and add equation (29) and (30),
(
)
β
2Ce = Ee
1+i
α
(
)
iβL
Ee
β
C =
1+i
.
2eαL
α
αL
iβL
(32)
Now subtract equation (30) from (29) we get,
(
)
β
2De
= Ee
1−i
α
(
)
iβL
β
Ee
1−i
D =
.
2e−αL
α
−αL
Date: 21 April, 2014
iβL
(33)
43
Quantum Mechanics I: PHY-212
Spring Semester 2014
Inserting values of C and D from equations (32) and (33) into equation (31) results in,
(
)(
)(
)
(
)(
)(
)
1 EeiβL
β
α
1 EeiβL
β
α
A =
1+i
1−i
+
1−i
1+i
2 2eαL
α
k
2 2e−αL
α
k
[
(
)
(
)]
iβL
Ee
β
α β
α
β β
e−αL 1 + i − i +
+ eαL 1 + i − i +
=
4
α
k
k
k
α k
[
{(
)
(
)}
{(
)
(
)}]
EeiβL −αL
β
β α
β
β α
αL
−
−
=
e
1+
+i
+e
1+
−i
4
k
α k
k
α k
[
(
)
(
)
(
)
(
)]
β
β
EeiβL −αL
α
α
−αL β
αL
αL β
e
1+
+ ie
+e
1+
− ie
.
=
−
−
4
k
α k
k
α k
and after some rearrangement,
[
(
)
(
)]
EeiβL
β
β α
αL
−αL
αL
−αL
A =
(e + e
) 1+
− i(e − e
)
−
4
k
α k
[( αL
)(
)
( αL
)(
)]
EeiβL
e + e−αL
β
e − e−αL
β α
=
1+
−i
.
−
2
2
k
2
α k
Using cosh x =
ex +e−x
2
and sinh x =
ex −e−x
,
2
[
(
)
(
)]
β α
EeiβL
β
A =
cosh(αL) 1 +
− i sinh(αL)
−
.
2
k
α k
Taking the complex conjugate of equation (34),
[
(
)
(
)]
β α
E ∗ e−iβL
β
∗
A =
cosh(αL) 1 +
+ i sinh(αL)
−
.
2
k
α k
(34)
(35)
Multiply equations (34) and (35) yields,
[
(
)
(
)]
EeiβL
β
β α
∗
AA =
cosh(αL) 1 +
− i sinh(αL)
−
·
2
k
α k
[
(
)
(
)]
E ∗ e−iβL
β
β α
cosh(αL) 1 +
+ i sinh(αL)
−
2
k
α k
(
(
[
)2
)2 ]
β α
|E|2
β
2
2
2
|A| =
cosh (αL) 1 +
+ sinh (αL)
−
4
k
α k
[
(
)2
(
)2 ]
1
|A|2
β
β α
2
2
=
−
(1 + sinh (αL)) 1 +
+ sinh (αL)
|E|2
4
k
α k
{(
[(
)2
)2 (
)2 }]
β
β α
1
β
2
+ sinh (αL) 1 +
+
−
=
1+
4
k
k
α k
1
= ·
T
If we would like to express the final result explicitly in terms of the energy and length, we
Date: 21 April, 2014
44
Quantum Mechanics I: PHY-212
Spring Semester 2014
obtain. Substitute values of k, α and β we get,
√
√
[(
(
){(
2m(E−W0 ) )2
2m(E−W0 ) )2
2
√
|A|
1
2 L
−1
~
~
√
√
= T =
1+
+ sinh
2m(V0 − E)
1+
2mE
2mE
|E|2
4
~
~
√
√ ~
( 2m(E−W0 )
2m(V0 −E) )2 }]
√ ~
+
− √~
2m(V0 −E)
~
2mE
~
√
√
[(
)2
)2
(
){(
√
1
E − W0
L
E − W0
2
1+
=
+ sinh
2m(V0 − E)
1+
4
E
~
E
√
(√
)2 }]
E − W0
V0 − E
+
·
−
V0 − E
E
(g) If the barrier acts like a 50:50 beamsplitter, the transmission probability will be 0.5.
Setting T = 0.5, in the last expression of part (e),
[
(
)]−1
√
V02
2 L
1+
sinh
2m(V0 − E)
4E(V0 − E)
~
)
(
√
V02
2 L
1+
sinh
2m(V0 − E)
4E(V0 − E)
~
)
(
√
V02
2 L
sinh
2m(V0 − E)
4E(V0 − E)
~
4E(V0 − E)
V02
√
4E(V0 − E)
V0
√
2 Eε
V0
=
1
2
= 2
= 1
)
L√
= sinh
2m(V0 − E)
~
(
)
L√
= sinh
2m(V0 − E)
~
(
)
L√
= sinh
2mε ,
~
(
2
where ε = V0 − E. The above expression is the required relation between E, V0 and L.
Answer 17:
From the figure we can see that the wavenumber for region 0 ≤ x ≤ L/2 is,
√
2mE
,
k=
~
while the wavenumber for region L/2 ≤ x ≤ L is,
√
2m(E − V0 )
′
k =
·
~
Since k ′ < k, λ > λ′ , option (d) is the correct answer. The wavelength is longer in the right
half of the potential well.
Date: 21 April, 2014
45
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 18:
Option (d) is the correct answer. From the figure, we observe that in region I the wavefunction is,
kI2 =
2m(2V0 )
4mV0
2mE
=
=
,
2
2
~
~
~2
and likewise for region II,
2
=
kII
2mE
2m(V0 )
2mV0
=
=
·
2
2
~
~
~2
Hence,
√
kI
= 2.
kII
Since k = p/~ = mv/~, we have
vI =
√
2vII .
Answer 19:
In the absence of electric field,
λ =
h
h
=√
, and speed is v.
p
2mE
In the presence of electric field, total energy of the electron changes to (E − eu(x)), resulting
in the modified wavelength,
λ′ =
h
h
=√
, and the changed speed is v ′ .
′
p
2m(E − eu(x))
Ratio between the momentums and speed is,
√
√
p
mv
v
2mE
E
=
= ′ =
=
·
′
′
p
mv
v
2m(E − eu(x))
(E − eu(x))
Hence (c) is the correct answer.
Answer 20:
The correct option is (a). The probability density Ψ∗ (x, t)Ψ(x, t) = e−i(kx−ωt) ei(kx−ωt)
= 1 is independent of position.
So the particle can be found, with equal likelihood,
anywhere along the x axis.
Date: 21 April, 2014
46
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 21:
π ~
The energy of electron is n2 2mL
2 and in the ground state E1 =
2 2
π 2 ~2
.
2mL2
the quantum dot, the electron can be excited to the level, Ephoton +
If a photon shines on
π 2 ~2
2mL2
. If the electron is
to remain confined,
π 2 ~2
< V0
2mL2
π 2 ~2
< V0 − Ephoton
2mL2
1
2mL2
>
2
2
π ~
V0 − Ephoton
√
π 2 ~2
L >
,
2m(V0 − Ephoton )
Ephoton +
which is the minimum diameter that the quantum dot must have.
Answer 22:
We are given that,
ψ(x) =
L2
1
·
+ (x − xo )2 /α2
At x = x0 ,
ψ(x = x0 ) =
1
·
L2
We can calculate (x − x0 ) where the amplitude is half of maximum, i.e., ψ(x) =
1
.
2L2
1
1
= 2
2
2L
L + (x − xo )2 /α2
(x − xo )2
2L2 = L2 +
α2
(x − xo )2
L2 =
α2
2 2
α L = (x − xo )2
⇒
αL = (x − xo )
∆x ∼ 2αL.
According to uncertainty principle,
∆p ≥
Date: 21 April, 2014
~
~
=
·
∆x
2αL
47
Quantum Mechanics I: PHY-212
Spring Semester 2014
Since we are talking about estimates, option (a) is the correct answer.
Answer 23:
(a)
Fig.(a) shows the energy landscape before a voltage V0 is applied. If the tip is given a
positive potential, it’s potential energy, as seen by the electrons decreases. This results in a
downwards bending of the energy profile.
Fig. (c): Tunneling into metal
φ
eVo
EF (bulk)
φ
Fig. (d): Tunneling into air
φ
eVo
EF (bulk)
φ
EF (tip)
EF (tip)
d
d
Bulk
Air
Tip
Bulk
Air
Tip
Now Fig. (c) and (d) show two scenarios; (c) is when V0 is small such that an electron in
the bulk metal can directly tunnel into the metal tip while keeping its energy constant. The
thick arrow shows a tunneling electron. Whereas in (d), the V0 is large, resulting in the
excessive sloping of the potential energy — to such a large extent, that EF (bulk) is higher in
energy than EF (tip) + ϕ. This would result in the electron being tunneled into air, instead
of the metal.
(b) For tunneling into the metal tip, the required condition is evident from Fig.(c),
EF (bulk) ≤ EF (tip) + ϕ·
(36)
We observe that,
EF (bulk) + ϕ = EF (tip) + ϕ + eV0
⇒ EF (bulk) = EF (tip) + eV0 ·
(37)
Comparing equations (42) and (37), we find out eV0 < ϕ is the condition for tunneling into
the metal tip. Likewise from Fig.(d), it is readily observable that eV0 > ϕ is the condition
for tunneling into air.
Date: 21 April, 2014
48
Quantum Mechanics I: PHY-212
Spring Semester 2014
(c) For tunneling into the metal tip, eV0 < ϕ. To find the threshold voltage we set,
V0 =
ϕ
= 4 V.
e
Hence maximum electric field ensuring tunneling into the tip is,
E =
V0
4
=
= 2 × 109 V/m .
−9
d
2 × 10
Answer 24:
For a half infinite well potential, the energy E inside the well is directly proportional to k 2 ,
(E = ~2 k 2 /2m) and the wave number k is k = 2π/λ. Therefore the maximum wavelength
inside the well corresponds to the ground state energy. If the ground state energy is less
than V0 , we will, for sure, have at least one bound state. The maximum possible wavelength
inside the well is simply equal to 2a, such that,
k=
2π
2π
π
=
= ·
λ
2a
a
The energy for this longest wavelength inside the well is,
~2 k 2
~2 π 2
h2 π 2
=
=
2m
2ma2
8π 2 ma2
2
h
=
·
8ma2
E =
For the existence of bound states it is necessary that V0 > E. It implies that there will be
no bound states if,
V0 <
h2
,
8ma2
which is the required result.
Answer 25:
E>Vo
I
II
Vo
x=0
Date: 21 April, 2014
49
Quantum Mechanics I: PHY-212
Spring Semester 2014
2mE
,
~2
2m(E − V0 )
ψII (x) = Ceik2 x , with k22 =
·
~2
ψI (x) = Aeik1 x + Be−ik1 x , with k12 =
The wavefunctions in region I,
The wavefunctions in region II,
and
At x = 0,
A+B = C
and ik1 (A − B) = ik2 C ,
(38)
(39)
which are derived from continuity of wavefunction and its derivative at the discontinuity
(x = 0).
From equation (38),
k1
(A − B) = C
k2
(40)
Comparing equation (38) and (40),
k1
A+B =
(A − B)
k2
(
)
(
)
k1
k1
A 1−
= −B 1 +
k2
k2
(
)
B
1 − (k1 /k2 )
= −
A
1 + (k1 /k2 )
)
(
k2 − k1
= −
k2 + k1
The reflection coefficient is given by |B|
,
|A|2
(
)2 (
)2
k2 − k1
|B|2
(k2 − k1 )
= −
=
·
|A|2
(k2 + k1 )
k2 + k1
2
Setting this to 0.5, we have,
)2
k2 − k1
k2 + k1
k2 − k1
k2 + k1
k2 − k1
At first consider
k2 + k1
k2 − k1
√
k2 (1 − 0.5)
k2
k1
Date: 21 April, 2014
(
= 0.5
√
= ± 0.5
(41)
√
= + 0.5
√
= 0.5 (k2 + k1 )
√
= k1 (1 + 0.5)
√
1 + 0.5
1.707
√
=
=
= 5.82 ,
0.293
1 − 0.5
50
Quantum Mechanics I: PHY-212
Spring Semester 2014
which is the required ratio. Note that since k2 > k1 , for 50% reflection (or 50% transmission),
the discontinuity must in fact be a depression instead of an elevation, as shown below.
E
Vo
x=0
If instead, we take the other possibility, i.e. the R.H.S in equation (41) is with a negative
sign,
k2 − k1
k2 + k1
k2 − k1
√
k2 (1 + 0.5)
k2
k1
√
= − 0.5
√
= − 0.5 (k2 + k1 )
√
= k1 (1 − 0.5)
√
1 − 0.5
0.293
√
=
=
= 0.17 .
1.707
1 + 0.5
In this second case, k2 < k1 , so the discontinuity is an elevation, as shown below.
E
Vo
x=0
Answer 26:
In thermionic emission, heating increases the thermal energy of the electrons.
These
electrons are raised from the filled to the unfilled levels. Some of these excited electrons
obtain enough energy to overcome the work function and can therefore be ejected into air.
Answer 27:
Option (d) is the correct answer. The quantum dot is given a variable positive potential
V0 . An electron added to the quantum dot raises its coulomb energy by e2 /2Cdot . Hence
Date: 21 April, 2014
51
Quantum Mechanics I: PHY-212
Spring Semester 2014
if energy is to be conserved and the electron transfer to the dot is to be favored, the
starting energy of the dot must be lower by e2 /2Cdot , so that pickup of an extra electron
is energetically permissible. If V0 = 0, the electron cannot tunnel into n = 7 as it will raise
the overall energy of the dot. If, however, V0 = e/2Cdot , n = 7 is lowered in energy by
e2 /2Cdot and electron tunneling to n = 7 becomes energetically permissible. The electron
tunnel! While keeping its total energy constant. The quantum dot is a receptacle lowering
its energy in anticipation of an incoming electron, which raises the energy back to the
original. Hence four peaks corresponding to tunneling to n = 7, 8, 9, 10 are observed.
Answer 28:
Option (c) is the correct answer. If v is to go up in region III, k must increase. Since
√
k ∝ E − V , the difference between E and V must be higher, which is the situation in
Fig.(c).
Answer 29:
In the refracting medium, the electric potential is positive and hence the potential energy
seen by the electron, V , is lower. This means that the difference E − V is larger, k is larger,
and hence the speed is slower, v2 < v1 . Hence from Bethe’s law, sin α < sin β, α < β. The
electron beam bends away from the normal. Hence option (c) is the correct answer.
Answer 30:
As we know that when a quantum object encounters a wider barrier, the tunneling transmission probability is lower. If the barrier is thinner the tunneling probability is higher. In the
given figure we can see that the barrier on the right is thinner. Therefore it is more probable to find the electron in region C as compared to A. Hence option (b) is the correct answer.
Answer 31:
One need to think carefully about this. Consider the accompanying figure.
Date: 21 April, 2014
52
Quantum Mechanics I: PHY-212
Spring Semester 2014
E
Vo
The potential depression V0 is large. Let’s find the reflection probability R. In region I,
ψI (x) = Aeik1 x + Be−ik1 x ,
and for region II,
ψII (x) = Ceik2 x ,
where k12 =
2mE
,
~
and k2 =
2m(E+V0 )
.
~
At the point of the precipice, x = 0, ψI (0) = ψII (0)
′
and ψI′ (0) = ψII
(0). So A + B = C and ik1 (A − B) = ik2 C. Eliminating C from these
equations,
A+B =
=
(
)
k1
B 1+
=
k2
B
=
A
R =
=
ik1 (A − B)
ik2
k1
k1
A− B
k2
k2
(
)
k1
A
−1
k2
k1 − k2
k1 + k2
|B|2 k1
|A|2 k1
(
)2
k1 − k2
.
k1 + k2
(
)2
If V0 is very large, k2 ≫ k1 , R becomes − kk22 = 1. Since R ≃ 1, T = 0. There is zero
probability for the particle to “fall over the edge” and enter region II. Hence option (a) is
the correct answer.
Answer 32:
Date: 21 April, 2014
53
Quantum Mechanics I: PHY-212
Spring Semester 2014
We are given that,
(
)
π 2 ~2 n2x n2y n2z
Enx ,ny ,nz =
+ 2 + 2
2m a2
b
c
2 2
π ~ 2
=
(nx + n2y + n2z ) since a = b = c.
2m
For ground state (nx , ny , nz ) = (1, 1, 1), energy of the ground state will be,
π 2 ~2 2
E(1, 1, 1) =
(1 + 12 + 12 )
2m
π 2 ~2
= 3
.
2m
Similarly for first excited state, (nx , ny , nz ) = (2, 1, 1), energy of the first excited state will
be,
π 2 ~2 2
(2 + 12 + 12 )
2m
π 2 ~2
.
= 6
2m
Energy difference for these two energy levels is,
E(2, 1, 1) =
π 2 ~2
π 2 ~2
−3
2m
2m
π 2 ~2
= 3
.
2m
E(2, 1, 1) − E(1, 1, 1) = 6
Hence option (c) is the correct answer.
Answer 33:
As we know that general equation of wave function is,
Ψ(x, t) = Aeikx−ωt .
Comparison of this equation with the given wave equation of for the free particle yields,
k = 2.5 × 1011 m−1
ω = 2.1 × 1013 s−1 .
Mass of the particle can be calculated by the dispersion relation,
k
m
k
⇒m = 2
ω
2.5 × 1011
=
(2.1 × 1013 )2
= 5.7 × 10−16 kg.
ω2 =
Date: 21 April, 2014
54
Quantum Mechanics I: PHY-212
Spring Semester 2014
Hence option (d) is the correct answer.
You don’t really need to remember the dispersion relationship. Look at the TDSE in the
absence of V :
TDSE : −
~2 d2
d
Ψ(x, t) = i~ Ψ(x, t).
2
2m dx
dt
Inserting the supplied wave function into the above, the relationship ω 2 = k/m automatically pops out. Students are tempted to use E = ~ω and E = p2 /2m. The former
relationship does not hold for all particles, it is specific in its meaning—it says that energy
of a photon E is related to the frequency of the electromagnetic wave associated with the
photon. Blindly using these relations is wrong!
Answer 34:
Since the length of infinite well is very large i.e. 30 cm, for a small amount of energy 1 eV,
the number of nodes will be very large i.e. 4.9 × 108 . Since the number of nodes is very
large, the waves are “squeezed” close together, the de Broglie wavelength is extremely small
obscuring chances of observing the quantum wave behavior at the classical macroscopic
scale. At such a high value of n, quantum effects are not visible. Another way of looking at
this is that the wave function is such that the probability of finding the electron becomes
equal everywhere, i.e. it imparting the electron a continuous quality rather than quantized.
All of this ties in well with Bohr’s corresponding principle.
Answer 35:
We are given that,
At t = 0
At any time t
Date: 21 April, 2014
)
1 (
Ψ(x, 0) = √
ψ1 (x) + 2ψ2 (x)
5L
)
1 (
Ψ(x, t) = √
ψ1 (x)e−iE1 t/~ + 2ψ2 (x)e−iE2 t/~ .
5L
55
Quantum Mechanics I: PHY-212
Spring Semester 2014
If E1 is the energy of ground state, energy of first excited state will be E2 = 4E1 .
)
1 (
⇒ Ψ(x, t) = √
ψ1 (x)e−iE1 t/~ + 2ψ2 (x)e−i4E1 t/~
5L
)
π~
1 (
At t =
ψ1 (x)e−iπ + 2ψ2 (x)e−i4π
Ψ(x, t) = √
E1
5L
)
1 (
= √
ψ1 (x)(−1) + 2ψ2 (x)(+1)
5L
)
1 (
= √
−ψ1 (x) + 2ψ2 (x) .
5L
Now we need to find out what −ψ1 (x) + 2ψ2 (x) looks like. The construction is seen in the
series of diagrams below.
ψ1(x)
0.2
1.0
-0.2
0.8
-0.4
A
0.4
0.6
0.8
-A
-0.6
Invert
0.6
0.4
1.0
−ψ1 (x)
-0.8
-1.0
0.2
0.2
0.4
0.6
0.8
1.0
−ψ1 (x)+ 2 ψ2 (x)
2
2 ψ2 (x)
Add
1
1
0.2
0.2
0.4
0.6
0.8
1.0
0.4
0.6
0.8
1.0
-1
-1
-2
-2
Hence option (a) is the correct answer.
Answer 36:
Option (d) is the correct answer. The wavefunction is zero at x ≥ b because of the infinite
potential and extends into the region x ≤ a. Furthermore, the value of k increases and
wavelength decreases as we go from x = a to x = b.
Answer 37:
Option (a) is the correct answer.
Date: 21 April, 2014
56
Quantum Mechanics I: PHY-212
Spring Semester 2014
Answer 38:
We are given the wavefunction of free particle,
Ψ(x, t) = A(eikx + e−ikx )e−i ~
Et
⇒ Ψ∗ (x, t) = A ∗ (e−ikx + e+ikx )ei ~
Et
Ψ∗ (x, t)Ψ(x, t) = A∗ (e−ikx + e+ikx ) · A(eikx + e−ikx )e−i ~ ei
Et
Et
~
= A2 (1 + e−2ikx + e2ikx + 1)
= A2 (2 + e−2ikx + e2ikx ).
Using cos x =
eix +e−ix
2
|Ψ2 (x, t)|2 = A2 (2 + 2 cos(2kx))
= A2 (2 + 2 cos(2kx))
= 2A2 (1 + cos(2kx))
= 2A2 · 2 cos2 (kx))
= 4A2 · cos2 (kx))
1. True since p(x) = |Ψ2 (x, t)|2 is independent of time.
2. False since p(x) depends on x and changes with x.
3. True because the forward and backward propagating waves have equal amplitudes
and the probability density does not change with time.
Answer 39:
According to uncertainty principle,
∆p∆x ≥
~
.
2
If particle moves in a circle of radius r and angular momentum L, then
L = pr
⇒ ∆L = ∆pr
∆L
⇒ ∆p =
r
and
∆x = r∆θ.
Date: 21 April, 2014
57
Quantum Mechanics I: PHY-212
Spring Semester 2014
Using these values of ∆p and ∆x in uncertainty relation,
∆L
~
· r∆θ ≥
r
2
~
∆L∆θ ≥ .
2
Hence option (a) is the correct answer.
Answer 40:
We are given that,
Tip-sample distance = α = 1 nm−1 = 1 × 109 m−1
Distance covered by tip = ∆L = 0.1 nm = 0.1 × 10−9 m.
Tunneling probability is,
Ti = e−2αL .
If tip moves closer to the surface by ∆L, final tunneling probability will become,
Tf = e−2α(L−∆L) .
ratio of tunneling probabilities is,
e−2α(L−∆L)
Tf
=
Ti
e−2αL
= e2α∆L
= e2×10
9 ×0.1×10−9
= e0.2
= 1.22.
Hence there is a 22% increase in the tunneling current and the correct answer is (b).
Answer 41:
(a) We are given that,
E = pc.
Date: 21 April, 2014
58
Quantum Mechanics I: PHY-212
Spring Semester 2014
Uncertainty in energy is,
∆E ≈ c∆p.
(42)
Uncertainty in momentum ∆p for a photon confined to the tiny box of size ∆x can be
calculated by using the uncertainty relation,
∆x ∆p ≥
~
,
2
where ~ is the reduced Planck’s constant.
⇒ ∆p ≥
~
.
2∆x
Using this value of ∆p in equation (42) yields,
(
)
~
∆E ∼ c
2∆x
c~
∆E ∼
2∆x
(b) We are given that,
∆E ∼ E
c~
⇒E=
2∆x
Effective mass can be calculated by using energy-mass relationship,
E = meff c2 ,
where meff is the effective mass of the photon.
E
c2
c~
1
=
· 2
2∆x c
~
=
.
2c∆x
⇒ meff =
(c) We are given that,
Mass of the star = M
Radius of the star = R
Let mass of the object = m,
Date: 21 April, 2014
59
Quantum Mechanics I: PHY-212
Spring Semester 2014
and let speed of the object to escape from the star’s gravitational pull is vesc . In order to
escape from the gravitational pull of star, kinetic energy of the object should be at least
equal to the gravitational potential energy.
K.E. = P.E.
1 2
mM
⇒ mvesc.
= G
2
R
2GM
2
vesc. =
√R
2GM
vesc. =
.
R
Which is the required result.
(d) Using the relation of vesc derived in part (c)
√
2GM
vesc =
,
R
where vesc = c,
√
2GM
R
2GM
.
c2 =
R
c =
Setting R = ∆x and M = meff ,
c2 =
=
=
∆x2 =
∆x =
(e)
2Gmeff.
∆x
2G
~
·
∆x 2c∆x
G~
c∆x2
G~
c3
√
G~
.
c3
√
∆x =
=
=
G~
3
√c
6.67 × 10−11 N m2 /kg2 × 1.06 × 10−34 Js
(3.0 × 108 ms−1 )3
√
2.62 × 10−70 m2
= 1.6 × 10−35 m.
Date: 21 April, 2014
60
Quantum Mechanics I: PHY-212
Spring Semester 2014
(f ) We are given that,
Diameter of photon = 2 fm
−15
m
 = 2 × 10
Diameter of photon
2 × 10−15 m

 =
1.6 × 10−35 m/Planck length
in Planck’s length

= 1.24 × 1020 Planck lengths.
Answer 42:
We are given that,
Barrier length = L = 35 fm = 35 × 10−15 m
Barrier height = (V0 − E) = 5 MeV = 5 × 106 × 1.6 × 10−19 J
= 8 × 10−13 J
Tunneling probability = T = e−2αL ,
√
2m(V0 − E)
where
α =
~
and mass of the nucleus = m = 4 × 1.67 × 10−27 kg = 6.68 × 10−27 kg.
In order to calculate probability of escape T , let’s first calculate α.
√
2m(V0 − E)
α =
~
√
2 × 6.68 × 10−27 kg × 5 × 106 × 1.6 × 10−19 J
=
1.06 × 10−34 Js
= 9.75 × 1014 m−1 .
Tunneling probability is,
T = e−2αL
= e−2×9.75×10
14
m−1 ×35×10−15 m
= e−68.25
= 2.29 × 10−30 .
Date: 21 April, 2014
61
Quantum Mechanics I: PHY-212
Spring Semester 2014
The probability that an alpha particle hitting the nucleus surface will escape is 2.29 × 10−30 .
If no. of hits per second = 5 × 1021 /s


Probability of escape

 = T = 2.29 × 10−30 × 5 × 1021 /s × 24 × 60 × 60 s
in a day
= 9.89 × 10−4
= 9.89 × 10−2 %
Date: 21 April, 2014
62